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3 years ago

Control variables are parts of the experiment that you try to:

Chemistry

2 answers:

Katena32 [7]3 years ago

6 0

Answer:

A control variable in scientific experimentation is an experimental element which is constant and unchanged throughout the course of the investigation.

Hope this helped :)

maks197457 [2]3 years ago

3 0

Answer:

study, you do not change them as they are meant to be controls not like independent variables.

Explanation:

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Which of the following is the definition of motion?

Irina-Kira [14]

Answer:

Explanation:

A change of position over time.

B is a vector and C is a distance.

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3 years ago

When Oxygen in the air reacts with Iron, Iron Oxide forms.

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Answer:

I think its E. iron and iron oxide have the same properties if it's not e its d for sure

Explanation:

iron + oxygen → iron oxide. 4 Fe + 3 O2 → 2 Fe2O.

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3 years ago

What discovery led J.J. Thomson to decide that atoms were not indivisible, but are actually composed of smaller parts?

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4 years ago

Imagine the reaction A + B ⇌ C + D is at equilibrium, where the forward and reverse reactions are at equal rates. What would hap

kondaur [170]

Answer:

Rate of forward reaction will increase.

Explanation:

Effect of change in reaction condition on equilibrium is explained by Le Chatelier's principle. According to this principle,

If an equilibrium condition of a dynamic reversible reaction is disturbed by changing concentration, temperature, pressure, volume, etc, then reaction will move will in a direction which counteract the change.

In the given reaction,

A + B ⇌ C + D

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As concentration A decreases in forward direction, therefore, rate of forward reaction will increase.

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4 years ago

A compound is 7.74% hydrogen and 92.26% carbon by mass. At 100°C a 0.6883 g sample of the gas occupies 250 mL when the pressure

ycow [4]

Answer: The molecular formula for the compound is $C_6H_6$

Explanation:

We are given:

Percentage of C = 92.26 %

Percentage of H = 7.74 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 92.26 g

Mass of H = 7.74 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.68 moles.

For Carbon = $\frac{7.68}{7.68}=1$

For Hydrogen = $\frac{7.74}{7.68}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is $CH$

Calculating the molar mass of the compound:

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{m}{M}RT$

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = $62.3637\text{ L. torr }mol^{-1}K^{-1}$

T = temperature of the gas = $100^oC=(100+273)K=373K$

Putting values in above equation, we get:

$820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol$